Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{r - 3}{-3r^2 + 30r - 63} \times \dfrac{r^2 - 5r - 14}{2r - 14} $
Answer: First factor out any common factors. $q = \dfrac{r - 3}{-3(r^2 - 10r + 21)} \times \dfrac{r^2 - 5r - 14}{2(r - 7)} $ Then factor the quadratic expressions. $q = \dfrac {r - 3} {-3(r - 3)(r - 7)} \times \dfrac {(r - 7)(r + 2)} {2(r - 7)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {(r - 3) \times (r - 7)(r + 2) } { -3(r - 3)(r - 7) \times 2(r - 7)} $ $q = \dfrac {(r - 7)(r + 2)(r - 3)} {-6(r - 3)(r - 7)(r - 7)} $ Notice that $(r - 3)$ and $(r - 7)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {(r - 7)(r + 2)\cancel{(r - 3)}} {-6\cancel{(r - 3)}(r - 7)(r - 7)} $ We are dividing by $r - 3$ , so $r - 3 \neq 0$ Therefore, $r \neq 3$ $q = \dfrac {\cancel{(r - 7)}(r + 2)\cancel{(r - 3)}} {-6\cancel{(r - 3)}(r - 7)\cancel{(r - 7)}} $ We are dividing by $r - 7$ , so $r - 7 \neq 0$ Therefore, $r \neq 7$ $q = \dfrac {r + 2} {-6(r - 7)} $ $ q = \dfrac{-(r + 2)}{6(r - 7)}; r \neq 3; r \neq 7 $